zoukankan html css js c++ java

HDOJ2222 Keywords Search[ac自动机模版题]

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19868 Accepted Submission(s): 6669

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

Output

Print how many keywords are contained in the description.

Sample Input

1 5 she he say shr her yasherhs

Sample Output

3

Author

Wiskey

Recommend

lcy

题意：给n个单词和一个字符串，询问该字符串中有多少个给出的单词。

code:

  1 #include<iostream>
  2 using namespace std;
  3 
  4 struct node
  5 {
  6     node *fail;
  7     node *next[26];
  8     int count;
  9     node()
 10     {
 11         fail=NULL;
 12         count=0;
 13         memset(next,NULL,sizeof(next));
 14     }
 15 }*q[500001];
 16 
 17 char keyword[55];
 18 char str[1000010];
 19 
 20 void insert(char *str,node *root)
 21 {
 22     node *p=root;
 23     int i=0;
 24     while(str[i])
 25     {
 26         int index=str[i]-'a';
 27         if(p->next[index]==NULL)
 28             p->next[index]=new node();
 29         p=p->next[index];
 30         i++;
 31     }
 32     p->count++;
 33 }
 34 
 35 void acbuild(node *root)               //构建失败指针
 36 {
 37     int i;
 38     int head=0,tail=0;
 39     root->fail=NULL;
 40     q[tail++]=root;
 41     while(head!=tail)
 42     {
 43         node *temp=q[head++];
 44         node *p=NULL;
 45         for(i=0;i<26;i++)
 46         {
 47             if(temp->next[i]!=NULL)
 48             {
 49                 if(temp==root)
 50                     temp->next[i]->fail=root;
 51                 else
 52                 {
 53                     p=temp->fail;
 54                     while(p!=NULL)
 55                     {
 56                         if(p->next[i]!=NULL)
 57                         {
 58                             temp->next[i]->fail=p->next[i];
 59                             break;
 60                         }
 61                         p=p->fail;
 62                     }
 63                     if(p==NULL)
 64                         temp->next[i]->fail=root;
 65                 }
 66                 q[tail++]=temp->next[i];
 67             }
 68         }
 69     }
 70 }
 71 
 72 int query(node *root)
 73 {
 74     int i=0,cnt=0;
 75     int index;
 76     node *p=root;
 77     while(str[i])
 78     {
 79         index=str[i]-'a';
 80         while(p!=root&&p->next[index]==NULL)
 81             p=p->fail;
 82         p=p->next[index];
 83         p=(p==NULL)?root:p; 
 84         node *temp=p;
 85         while(temp!=root&&temp->count!=-1)
 86         {
 87             cnt+=temp->count;
 88             temp->count=-1;
 89             temp=temp->fail;
 90         }
 91         i++;
 92     }
 93     return cnt;
 94 }
 95 
 96 int main()
 97 {
 98     int n,t;
 99     scanf("%d",&t);
100     while(t--)
101     {
102         node *root=new node();
103         scanf("%d",&n);
104         getchar();
105         for(int i=0;i<n;i++)
106         {
107             gets(keyword);
108             insert(keyword,root);
109         }
110         acbuild(root);
111         scanf("%s",str);
112         printf("%d\n",query(root));
113     }
114     return 0;
115 }
116 /*
117 10
118 3
119 a
120 bb
121 ca
122 aabbcca
123 */

查看全文

相关阅读:
Android Permission 访问权限大全（转）
.NET中DateTime.Now.ToString的格式化字符串
 linux
code only
常用JavaScript操作页面元素的方法
 C#将字符串数组转换为以逗号分隔的字符串
 C#去除数组空格
 追源索骥：透过源码看懂Flink核心框架的执行流程
 高并发请求的缓存设计策略
 spark 2.3 导致driver OOM的一个SparkPlanGraphWrapper源码的bug

原文地址：https://www.cnblogs.com/XBWer/p/2686088.html